Question:

The one that can exhibit highest paramagnetic behaviour among the following is :
gly = glycinato; bpy = 2, 2'-bipyridine

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Find the oxidation state (ox is 2-), the d-electron count and the number of unpaired electrons in each complex.
Updated On: Nov 5, 2026
  • \(\mathrm{[Fe(en)(bpy)(NH_3)_2]^{2+}}\)
  • \(\mathrm{[Pd(gly)_2]}\)
  • \(\mathrm{[Co(OX)_2(OH)_2]^-}\ (\Delta_0 \gt P)\)
  • \(\mathrm{[Ti(NH_3)_6]^{3+}}\)
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The Correct Option is C

Solution and Explanation

(1) \(\mathrm{[Fe(en)(bpy)(NH_3)_2]^{2+}}\): Fe2+, \(d^6\); all ligands are N-donor strong-field (en, bpy, \(\mathrm{NH_3}\)), low spin \(t_{2g}^6\): 0 unpaired electrons.
(2) \(\mathrm{[Pd(gly)_2]}\): Pd2+, \(d^8\), square planar: 0 unpaired electrons.
(3) \(\mathrm{[Co(ox)_2(OH)_2]^-}\): ox is \(2-\) each and OH is \(1-\) each, so \(x - 4 - 2 = -1\), \(x = +5\): Co5+, \(d^4\). With \(\Delta_0 \gt P\) it is low spin, \(t_{2g}^4 e_g^0\): 2 unpaired electrons.
(4) \(\mathrm{[Ti(NH_3)_6]^{3+}}\): Ti3+, \(d^1\): 1 unpaired electron.
The maximum number of unpaired electrons (2) is in (3).
Answer: (3) \(\mathrm{[Co(OX)_2(OH)_2]^-}\)
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